Question

Let $f_k\left(x\right)=\frac{1}{k}({\sin }^{k}x+{\cos }^{k}x)$ where $xϵR$ and $k\ge 1$. Then $f_4\left(x\right)-f_6\left(x\right)$ equals

• A. $\frac{1}{12}$
• B. $\frac{1}{6}$
• C. $\frac{1}{3}$
• D. $\frac{1}{4}$

Answer 1

The correct option is A $\frac{1}{12}$

Given : $f_k\left(x\right)=\frac{1}{k}({\sin }^{k}x+{\cos }^{k}x)$

For $k=4$

$f_4\left(x\right)=\frac{1}{4}({\sin }^{4}x+{\cos }^{4}x)$
$=\frac{1}{4}({\left({\sin }^{2}x+{\cos }^{2}x\right)}^2-2{\cos }^{2}x{\sin }^{2}x)$
$(\because a^4+b^4=(a^2+b^2{)}^2-2a^2{b}^2,si{n}^{2}x+{\cos }^{2}x=1)$
$=\frac{1}{4}(1-2{\cos }^{2}x{\sin }^{2}x)$
For $k=6$
$f_6\left(x\right)=\frac{1}{6}({\sin }^{6}x+{\cos }^{6}x)$
$=\frac{1}{6}({\left({\sin }^{2}x+{\cos }^{2}x\right)}^3-3{\cos }^{2}x{\sin }^{2}x\left({\sin }^{2}x+{\cos }^{2}x\right))$
$(\because a^6+b^6=(a^2+b^2{)}^3-3a^2{b}^2(a^2+b^2),si{n}^{2}x+{\cos }^{2}x=1)$
$=\frac{1}{6}(1-3{\cos }^{2}x{\sin }^{2}x)$
Now $f_4\left(x\right)-f_6\left(x\right)$
$=\frac{1}{4}-\frac{1}{2}{\cos }^{2}x{\sin }^{2}x-\frac{1}{6}+\frac{1}{2}{\cos }^{2}x{\sin }^{2}x=\frac{1}{12}$